Enter the circle of radius $r$ centered at a point $P$, the set of all points the same distance $r$ away from $P$. Our very first objective is to find a parametric representation for the circle that we can do calculus with. While there is the familiar Cartesian representation of a circle centered at the origin:
$$x^2 + y^2 = r^2$$Which follows directly from the Pythagorean Theorem, as any point $(x, y)$ on the circle forms a right triangle with a radius as the hypotenuse, and which covers points with negative coordinates because squaring makes everything positive anyways. This representation will unfortunately prove problematic for reasons which will be discussed later. Instead, we will try to establish a parametric representation for the circle. Fortunately, there is a representation of this kind which is very nice and will prove very helpful:
Proposition I: The circle of radius $r$ centered at the origin is given by the rational parametrization:
$$x = r \cdot \frac{r^2 - t^2}{r^2 + t^2}$$ $$y = r \cdot \frac{2rt}{r^2 + t^2}$$Where $t$ is the $y$-coordinate of the point where a chord from the point $(-r, 0)$ to some point on the unit circle crosses the $y$-axis.
Proof: Let O be the origin. Draw the chord from the point P $= (-r, 0)$ to any point C on the circle. Let T be the point where this chord PC crosses the y-axis. Drop the altitude CX perpendicular to the $x$-axis. Observe that the triangles PTO and PCX are similar triangles. That is, they form the same slope. This means there must exist some scaling factor $s$ from PTO to PCX.
Let $t$ be the $y$-coordinate of T. Notice that OCX is a triangle whose hypotenuse is a radius of the circle. Since CX is perpendicular to the $x$-axis, this must be a right triangle. Since this triangle shares the side CX with PCX, which is similar to PTO, the Pythagorean Theorem applied to OXC tells us that:
$$\begin{align*} (st)^2 &= r^2 - (rs - r)^2\\ &= rs(2r - rs)\\ &= r^2s(2-s) \end{align*}$$Since we have a difference of squares. After more manipulation, this becomes:
$$\begin{align*} & st^2/r^2 = 2 - s\\ \Rightarrow\, & s(t^2 + r^2)/r^2 = 2\\ \Rightarrow\, & s = 2r^2/(t^2 + r^2) \end{align*}$$Since the length of CX is the magnitude of the $y$-coordinate corresponsing to $t$ in the parametrization, and the length of OX is the magnitude of the x coordinate of the parametrization, we get that:
$$y = st$$ $$x = sr - r$$From which the final parametrization follows from straightforward manipulation. $\square$
In the case of the unit circle, this parametrization reduces to:
$$x = \frac{1 - t^2}{1 + t^2}$$ $$y = \frac{2t}{1 + t^2}$$The scaling factor between the two similar triangles turns out to be pretty important. We will denote it as $s_r (t)$ for a radius $r$.
$$s = \frac{2r^2}{r^2 + t^2}$$With a parametrization for the circle under our belt, our next endeavor shall be to find the circumference of the circle and its area. The former is easy once the underlying work of defining integration and arclength has been done. Recall that the arclength of a parametric curve is given by the integral:
$$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$In order to make use of this arclength formula, we first need to take derivatives of each of the functions in the parametrization. Fortunately rational functions don't require any fancy tricks. Despite this, the result of the calculation turns out to be rather interesting.
Proposition II: The derivatives of the functions defining the parametrization of a circle with radius $r$ are given by:
$$\frac{dx}{dt} = - \frac{1}{r} sy$$ $$\frac{dy}{dt} = \frac{1}{r} sx$$Now we are finally ready to prove an important theorem.
Proposition III: The ratio of a circle's circumference to its diameter is always the same.
Proof: Since the parametrization of the unit circle must satisfy:
$$x^2 + y^2 = r^2$$We have that:
$$\begin{align*} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} &= \sqrt{\frac{1}{r^2} s^2y^2 + \frac{1}{r^2} s^2\,x^2}\\ &= \frac{1}{r} s \sqrt{y^2 + x^2}\\ &= \frac{1}{r} s \sqrt{r^2}\\ &= s \end{align*}$$Which is what we want to integrate to find the arc length. But what interval are we going to integrate over? The question is equivalent to figuring out which two chords we can slice the circle with that will account for the whole circumference. Unfortunately, there are no such chords, as accounting for the whole circumference would essentially require a tangent line going through $(-r, 0)$. Fortunately, we can get as close to that tangent line as we want by selecting larger and larger values for the y-coordinates our chords will cross. Thus, we will want to integrate over the entire real line.
By using the substitution $ru= t \Rightarrow r\,du = dt$, the circumference calculation becomes:
$$\begin{align*} C &= \int_{-\infty}^{\infty} s\, dt\\ &= 2r \int_{-\infty}^{\infty} \frac{du}{1 + u^2} \end{align*}$$Which shows that the ratio between the circumference and the diameter of any circle will always come out to be the same integral, as desired. $\square$
The above discovery motivates us to make the following definition:
Definition: The number $\pi$ is the ratio of a circle's circumference to its diameter. This is equivalent to:
$$\pi = \int_{-\infty}^{\infty} \frac{du}{1 + u^2}$$Furthermore, we also have that the circumference of a circle is given by:
$$C = 2\pi r$$Proposition IV: The area of a circular sector of radius $r$ subtended by an arc of length $\Theta$ is given by $\frac{1}{2}r \Theta$.
Proof: Let $\Theta$ be the arclength from the point at $t = a$ to $t = b$ for some real $b > a$. Then, by the arclength formula, we have: $$\Theta = \int_{a}^{b} s\, dt$$ We can use the parametric area formula to compute the area of the circular sector over this same interval of values: $$\begin{align*} A &= \frac{1}{2} \int_{a}^{b} x\frac{dy}{dt} - y\frac{dx}{dt} \,dt\\ &= \frac{1}{2} \int_{a}^{b} \frac{1}{r}sx^2 + \frac{1}{r}sy^2\,dt\\ &= \frac{1}{2} \int_{a}^{b} \frac{1}{r} sr^2\, dt\\ &= \frac{1}{2} r \int_{a}^{b} s\, dt\\ &= \frac{1}{2} r\Theta \end{align*}$$
Which is the desired result. $\square$
The formula for the area of the circle then follows as a corollary by integrating over the entire real line.
Exercise I: Complete the proof of Proposition I.
Exercise II: Prove Proposition II. Hint: It is easiest to start with the case r = 1, then generalize.
Exercise III: Show that the integral $\int_{-\infty}^{\infty} \frac{du}{1 + u^2}$ is less than or equal to 4.